Equations P $UnitSystem K kpa. F luid$ = Air (1) Input data for fluid. $If Fluid$= Air. C P = [kj/kg K] ; k = 1.

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P09-169 Equations Thermodynamics - An Engineering Approach (5th Ed) - Cengel, Boles - Mcgraw-Hill (2006) - pg.

1 P Equations Thermodynamics - An Engineering Approach (5th Ed) - Cengel, Boles - Mcgraw-Hill (2006) - pg. 549 Ciclul Brayton cu regenerare (aer, heliu) cu comprimare si destindere in trepte, c p =ct. Sa se determine efectul numarului de trepte de comprimare si destindere asupra randamentului unui ciclu Brayton regenerativ ideal cu comprimare si destindere in trepte. Raportul total de comprimare e 12, iar aerul intra in fiecare treapta a compresorului cu 300K si in fiecare treapta a turbinei cu 1200 K. Daca pentru aer se considera calduri specifice constante la temperatura ambianta, sa se determine randamentul termic al ciclului prin varierea numarului de trepte. Sa se compare rezultatele obtinute cu randamentul unui ciclu Ericsson care functioneaza intre aceleasi limite de temperatura. $UnitSystem K kpa F luid$ = Air (1) Input data for fluid $If Fluid$= Air C P = [kj/kg K] ; k = 1.4; (2) $Else C P = [kj/kg K] ; k = 1.667; (3) $EndIf 1

2 Marimi de intrare: N stages = 1 Nstages is the number of compression and expansion stages (4) T 6 = 1200 [K] (5) P ratio = 12 T 1 = 300 [K] (6) (7) P 1 = 100 [kpa] (8) η reg = 1.0 regenerator effectiveness (9) η c = 1.0 Compressor isentorpic efficiency (10) η t = 1.0 Turbine isentropic efficiency (11) Rezolvare: R p = P 1/Nstages ratio (12) Isentropic Compressor anaysis T 2s = T 1 R k 1 k p [K] (13) P 2 = R p P 1 [kpa] (14) T 2s is the isentropic value of T 2 at compressor exit η c = w comp,isen /w comp (15) compressor adiabatic efficiency, W comp > w comp,isen Conservation of energy for the compressor for the isentropic case: e in - e out = e=0 for steady-flow w comp,isen = C P (T 2s T 1 ) [kj/kg] (16) Actual compressor analysis: w comp = C P (T 2 T 1 ) [kj/kg] (17) Since intercooling is assumed to occur such that T 3 = T 1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is: w comp,total = N stages w comp [kj/kg] (18) 2

3 External heat exchanger analysis SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e in - e out = e cv =0 for steady flow The heat added in the external heat exchanger + the reheat between turbines is q in,total = C P (T 6 T 5 ) + (N stages 1) C P (T 8 T 7 ) [kj/kg] (19) Reheat is assumed to occur until: T 8 = T 6 [K] (20) Turbine analysis P 7 = P 6 R p [kpa] (21) T 7s is the isentropic value of T 7 at turbine exit T 7s = T 6 (1/R p ) k 1 k [K] (22) Turbine adiabatic efficiency, w turbisen > w turb η t = w turb /w turb,isen (23) SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e in -e out = e cv = 0 for steady-flow w turb,isen = C P (T 6 T 7s ) [kj/kg] (24) Actual Turbine analysis: w turb = C P (T 6 T 7 ) [kj/kg] (25) w turb,total = N stages w turb [kj/kg] (26) Cycle analysis w net = w turb,total w comp,total [kj/kg] (27) Bwr = w comp /w turb Back work ratio (28) P 4 = P 2 ; P 5 = P 4 ; P 6 = P 5 ; T 4 = T 2 ; (29) The regenerator effectiveness gives T 5 as: η reg = T 5 T 4 T 9 T 4 (30) 3

4 T 9 = T 7 [K] (31) Energy balance on regenerator gives T 10 as: T 4 + T 9 = T 5 + T 10 (32) Cycle thermal efficiency with regenerator η th,regenerative = w net /q in,total 100 [%] (33) The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T 6 and T 1 for this problem. η th,ericsson = (1 T 1 /T 6 ) 100 [%] (34) Data$ = Date$; (35) Solution Bwr = C P = [kj/kg-k] Data$ = η c = 1 η reg = 1 η t = 1 η th,ericsson = 75 η th,regenerative = F luid$ = Air k = 1.4 N stages = 1 P 1 = 100 [kpa] P 2 = 1200 [kpa] P 4 = 1200 [kpa] P 5 = 1200 [kpa] P 6 = 1200 [kpa] P 7 = 100 [kpa] P ratio = 12 q in,total = [kj/kg] R p = 12 T 1 = 300 [K] T 10 = [K] T 2 = [K] T 2s = [K] T 4 = [K] T 5 = 590 [K] T 6 = 1200 [K] T 7 = 590 [K] T 7s = 590 [K] T 8 = 1200 [K] T 9 = 590 [K] w comp = [kj/kg] w comp,isen = [kj/kg] w comp,total = [kj/kg] w net = [kj/kg] w turb = [kj/kg] w turb,isen = [kj/kg] w turb,total = [kj/kg] Table 1 Run F luid$ N stages η th,regenerative η th,ericsson

5 Thermal Efficiency vs Number of Stages of Intercooling and Reheat 5

$IfNot ParametricTable= P_ratio_gas. P ratio,gas = 14; Raport comprimare compresor aer - Pressure ratio for gas compressor (2) $EndIf

$IfNot ParametricTable= P_ratio_gas. P ratio,gas = 14; Raport comprimare compresor aer - Pressure ratio for gas compressor (2) $EndIf P10-078 Equations Thermodynamics - An Engineering Approach (5th Ed) - Cengel, Boles - Mcgraw-Hill (2006) - pg. 598 Centrala cu cicluri combinate Se considera o centrala electrica cu ciclu combinat gaze-abur